Let $f$ be a vector-valued function defined by $f(t)=\left(\sqrt{t+4},\dfrac6{t+2}\right)$. Find $f$ 's second derivative $f''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-\dfrac{1}{4\sqrt{(t+4)^3}},\dfrac{12}{(t+2)^3}\right)$ (Choice B) B $\left(-\dfrac{2}{\sqrt{(t+4)}},-\dfrac{12}{(t+2)}\right)$ (Choice C) C $\left(\sqrt{(t+4)^3},\dfrac{24}{(t+2)^2}\right)$ (Choice D) D $\left(\dfrac{1}{2\sqrt{(t+4)^3}},-\dfrac{6}{(t+2)^2}\right)$
Solution: We are asked to find the second derivative of $f$. This means we need to differentiate $f$ twice. In other words, we differentiate $f$ once to find $f'$, and then differentiate $f'$ (which is a vector-valued function as well) to find $f''$. Recall that $f(t)=\left(\sqrt{t+4},\dfrac6{t+2}\right)$. Therefore, $f'(t)=\left(\dfrac{1}{2\sqrt{t+4}},-\dfrac{6}{(t+2)^2}\right)$. Now let's differentiate $f'(t)=\left(\dfrac{1}{2\sqrt{t+4}},-\dfrac{6}{(t+2)^2}\right)$ to find $f''$. $f''(t)=\left(-\dfrac{1}{4\sqrt{(t+4)^3}},\dfrac{12}{(t+2)^3}\right)$ In conclusion, $f''(t)=\left(-\dfrac{1}{4\sqrt{(t+4)^3}},\dfrac{12}{(t+2)^3}\right)$.